Cho hàm số f(x) thỏa mãn [f'(x)]^2+ f(x)f''(2x)= 15x^4- 36x^2+6x

* Hướng dẫn giải

Chọn đáp án A

Theo đề bài: ${\left[f\text{'}\left(x\right)\right]}^2+f\left(x\right).f\text{'}\text{'}\left(x\right)=15x^4-36x^2+6x+9$

$\begin{array}{l}\Rightarrow \left[f\left(x\right).f\text{'}\left(x\right)\right]\text{'}=15x^4-36x^2+6x+9\\ \Rightarrow f\left(x\right).f\text{'}\left(x\right)=\int \left(15x^4-36x^2+6x+9\right)dx=3x^5-12x^3+3x^2+9x+C\end{array}$

Mà $f\left(0\right).f\text{'}\left(0\right)=-3\Rightarrow C=-3$

Do đó: $f\left(x\right).f\text{'}\left(x\right)=3x^5-12x^3+3x^2+9x-3$

$\begin{array}{l}\Rightarrow \underset{0}{\overset{1}{\int }}f\left(x\right).f\text{'}\left(x\right)dx=\underset{0}{\overset{1}{\int }}\left(3x^5-12x^3+3x^2+9x-3\right)dx\\ \Rightarrow \frac{1}{2}{\left.f^2\left(x\right)\right|}_0^1=0\Rightarrow f^2\left(1\right)-f^2\left(0\right)=0\\ \Rightarrow f^2\left(1\right)=1.\end{array}$